By William F. Stout

ISBN-10: 0126727503

ISBN-13: 9780126727500

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**Extra resources for Almost Sure Convergence**

**Example text**

1) There are three cases to consider. (i) Assume r 2 = N + 1 (no upcrossings). In this case, dN = 0 and /Ly = 0, establishing Eq. 1). (ii) Assume hN > 0 and n odd. ) In this case dN < (a — b)hN, using the fact that the number of upcrossings hN equals the number of downcrossings after the first upcrossing and that the length of each down crossing is less than or equal to a — b. Thus Eq. 1) holds. (iii) Assume hN > 0 and n even (last crossing an upcrossing). In this case dN < (a — b){hN — 1) + uN — utn, using the fact that the number of downcrossings after the first upcrossing equals the number of upcrossings minus one, that is, hN — 1.

Let Un = Ya=i Yt for each n > 1 define the martingale {Un, n > 1}. Let t be the first integer n > 1 such that Un > 1. Clearly EUt = 1 yet EUX = 0. Interestingly EUil) = 0 for all n > 1. 2. s. for some M < oo. Prove that E t Yt = 0. ) | Besides the optional stopping discussed above, optional sampling of a martingale (or submartingale) is possible: Let 1 < tx < t2 < • • • < tn < • • • be a sequence of stopping rules and {Un, n > 1} a martingale. Does it follow that Utl, Ut2, . . , Utn, . . is also a martingale?

The statistical problem is to decide whether {pn, n > 1} or {qn, n > 1} is the correct set of densities. 5. for each n > 1. The Tn are known as likelihood ratios. If the pn are correct, it seems plausible that the observed Yh will, when substituted into Tn, make Tn approach zero as n gets large. Likewise, if the qn are correct, it seems plausible that the observed Yh will make Tn approach infinity as n gets large. In this way, Tn, for large n, can be observed to decide whether the pn or the qn seem to be the correct densities.

### Almost Sure Convergence by William F. Stout

by James

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